Help Wanted with (Probably) New Sequence

[David Harden]

I need help extending a sequence I have in mind from those who are familiar with the work on primitive permutation groups.

The definition is this:

This is the set of all n such that for all finite groups G and all primes p, the number of Sylow p-subgroups of G does not equal n.

Here are the first 4 terms: 2, 22, 34, 46

Here is a proof of that:
Recall:
(i) If n is an odd positive integer, then the number of Sylow 2-subgroups of the dihedral group of order 2n is n. Therefore all terms of this sequence are even.
(ii) Let p be a prime, and let q be a power of p. The number of Sylow p-subgroups of PGL(2,q) is q+1.
(iii) Let q be a power of a prime, and let r be a prime factor of q-1. Then the group of all affine transformations from the finite field F_q to itself has exactly q Sylow r-subgroups. This implies that q is a nonmember of the sequence, as long as q-1 has prime factors. This works for all q except 2.
(iv) The Third Sylow Theorem: If G is a finite group and p is a prime factor of |G|, then the number of Sylow p-subgroups of G is congruent to 1 mod p.

2 is in the sequence, since 2 is not congruent to 1 modulo any primes. It is never again this easy. Therefore we need an easier equivalent form of the membership criterion:

Easier Test. n > 2 is in this sequence iff the symmetric group S_n lacks a transitive subgroup with exactly n Sylow p-subgroups.

Proof. If S_n has a transitive subgroup with exactly n Sylow p-subgroups, then n trivially is a nonmember of this sequence.

For the other direction, suppose that G has exactly n Sylow p-subgroups. Then we will show that S_n has a transitive subgroup with exactly n Sylow p-subgroups.
Let P be a Sylow p-subgroup of G and let N be its normalizer in G. Let G act on its Sylow p-subgroups by conjugation. This gives a homomorphism phi: G --> S_n. We want to show that phi(G) is also a group with exactly n Sylow p-subgroups. It is transitive by the Second Sylow Theorem.
First, phi(P) is a nontrivial subgroup of G: By properties of Sylow subgroups, P normalizes none of the other Sylow p-subgroups of G. Therefore phi(P) is a p-subgroup of G. Also, [phi(G):phi(P)] divides [G:P], which is a nonmultiple of p. Therefore phi(P) is a Sylow p-subgroup of phi(G).
Since N normalizes P, phi(N) normalizes phi(P).
Finally, nothing outside phi(N) can normalize phi(P), by the definition of phi. Likewise phi(N) is a point-stabilizer in phi(G).
Since phi(G) is a transitive subgroup of S_n, the index of phi(N) in phi(G) is n. Since phi(N) is the normalizer of phi(P) in phi(G), this means the number of Sylow p-subgroups of phi(G) is n, and we are done.

We continue the notation used above with G, N and P in what follows.

22 is in the sequence:
If G is a finite group with exactly 22 Sylow p-subgroups, then p= 3 or 7. (Since there is no need to divide into cases based on the value of p, in what follows p= 3 or 7.)
22 has no proper divisors congruent to 1 mod p, so N is a maximal subgroup of G. Then the Easier Test can be strengthened to say "primitive" where it says "transitive".
There are only 4 nonisomorphic primitive subgroups of S_22: M_22, Aut(M_22), A_22 and S_22. All of these have more than 22 Sylow p-subgroups.

34 is in the sequence:
If G is a finite group with exactly 34 Sylow p-subgroups, then p= 3 or 11.
34 has no proper divisors congruent to 1 mod p, and (proceeding as before) the only primitive subgroups of S_34 are S_34 and A_34. Both have more than 34 Sylow p-subgroups.

46 is in the sequence:
If G is a finite group with exactly 46 Sylow p-subgroups, then p= 3 or 5.
46 has no proper divisors congruent to 1 mod p, and the only primitive subgroups of S_46 are S_46 and A_46.

Every other even positive integer < 56 is a nonmember of the sequence:
Most of these are 1 more than some power of a prime and so are covered by (ii). 16 is covered by (iii). 36 is the number of Sylow 5-subgroups of A_5 x A_5. 40 is the number of Sylow 3-subgroups of A_4 x A_5. 52 is the number of Sylow 3-subgroups of A_4 x <x,y| x^13 = y^12 = 1, yxy^(-1) = x^2>.

I do not know about the membership status of 56, since there are 9 primitive permutation groups of degree 56 (according to OEoIS) and I know only 4 of them (S_56, A_56, S_8 and A_8).

(What follows are condensed arguments, intended to be filled in with the logic used above.)
58 is a member, since the only primitive permutation groups of degree 58 are S_58 and A_58.

(some cases in between: 66 excluded by G=PSL(2,11), p=5, 70 excluded by G=S_7, p=3, 76 excluded by G=A_4 x <x,y|x^19=y^3=1, yxy^(-1)=x^7>, p=3, 78 excluded by G=PSL(2,13), p=7)

The next member is 86: the only primitive permutation groups of degree 86 are S_86 and A_86.

The next member is 88: the only primitive permutation groups of degree 88 are S_88 and A_88.
Then an extension K of a Sylow 3-subgroup normalizer N must have [K:N]=4 (22 was already ruled out). K will be maximal in G and have [G:K]=22. None of M_22, Aut(M_22), A_22 or S_22 has exactly 88 Sylow 3-subgroups.

The next member is 92: the only primitive permutation groups of degree 92 are S_92 and A_92.

The next member is 94: the only primitive permutation groups of degree 94 are S_94 and A_94.

(some in between: 96 excluded by G=A_5 x (affine group of finite field of order 16), p=5, 100 excluded by G=A_5 x A_5, p=3,)

The next member is 106: the only primitive permutation groups of degree 106 are S_106 and A_106.

(some in between: 112 excluded by A_4 x PSL(2,27))

The next member is 116: the only primitive permutation groups of degree 116 are S_116 and A_116.

The next member is 118: the only primitive permutation groups of degree 118 are S_118 and A_118.

(some in between: 120 excluded by G= S_7, p=7)

The next member is 134: the only primitive permutation groups of degree 134 are S_134 and A_134.

(some in between: 136 excluded by G=PSL(2,16), p=5)

The next member is 142: the only primitive permutation groups of degree 142 are S_142 and A_142.

The next member is 146: the only primitive permutation groups of degree 146 are S_146 and A_146.

The next member is 154: the only primitive permutation groups of degree 154 are S_154 and A_154. All of M_22, Aut(M_22), A_22 and S_22 have more than 154 Sylow 3-subgroups. (see the reasoning with 88)

The next number with status I do not know is 162, since I do not know all the primitive permutation groups of degree 162 and how many Sylow 7-subgroups or Sylow 23-subgroups they have.

The next member is 166: the only primitive permutation groups of degree 166 are S_166 and A_166.

The next member is 178: the only primitive permutation groups of degree 178 are S_178 and A_178.

The next member is 184: the only primitive permutation groups of degree 184 are S_184 and A_184. The only primitive permutation groups of degree 46 are S_46 and A_46, and they do not work either. (this follows the 88 reasoning)

The next member is 188: the only primitive permutation groups of degree 188 are S_188 and A_188.

The next member is 202: the only primitive permutation groups of degree 202 are S_202 and A_202.

The next member is 204: the only primitive permutation groups of degree 204 are S_204 and A_204.

The next member is 206: the only primitive permutation groups of degree 206 are S_206 and A_206.

208 is the number of Sylow 3-subgroups of A_4 x A_4 x <x,y| x^13=y^3=1, yxy^(-1) = x^3>.

I don't know all the primitive permutation groups of degree 210 and how many Sylow 11-subgroups or Sylow 19-subgroups they have. This is where I will stop in this email. (I apologize if this length is inconvenient, and I apologize if the inconsistencies in the depth of treatment are irritating.)

---- David